On certain sufficient conditions for starlikeness

Vol. 23, No. 8 (2000) 521–527 S0161171200003574 ©Hindawi Publishing Corp.

ON CERTAIN SUFFICIENT CONDITIONS FOR STARLIKENESS

NIKOLA TUNESKI

(Received 22 October 1998 and in revised form 29 June 1999)

Abstract.We consider certain properties off (z)f_(z)/f2_{(z)as a sufficient condition} for starlikeness.

Keywords and phrases. Univalent, starlike.

2000 Mathematics Subject Classification. Primary 30C45.

1. Introduction and preliminaries. LetAdenote the class of functionsf (z)which are analytic in the unit discU= {z:|z|<1}withf (0)=f_(0)−1=0.

For a functionf (z)∈Awe say that it isstarlikein the unit discUif and only if

Rezf(z) f (z)

>0 (1.1)

for allz∈U. We denote byS∗_{the class of all such functions. We denote byKthe class} ofconvex functions in the unit discU, i.e., the class of univalent functionsf (z)∈A for which

Re1+zf(z) f_(z)

>0, (1.2)

for allz∈U.

Both of the above mentioned classes are subclasses of univalent functions inUand moreK⊂S∗_{([1, 2]).}

Letf (z)andg(z)be analytic in the unit disc. Then we say thatf (z)issubordinateto g(z), and we writef (z)≺g(z), ifg(z)is univalent inU,f (0)=g(0)andf (U)⊆g(U). In this paper, we use the method of differential subordinations. The general theory of differential subordinations introduced by Miler and Mocanu is given in [5]. Namely, ifφ:C2_{→C(whereCis the complex plane) is analytic in domainD, ifh(z)is univalent} inU, and ifp(z)is analytic inUwith(p(z),zp_{(z))∈Dwhenz∈U, then we say that}

p(z)satisfies a first-order differential subordination if

φp(z),zp_{(z)≺h(z). (1.3)}

We say that the univalent functionq(z)isdominantof the differential subordination (1.3) ifp(z)≺q(z) for allp(z) satisfying (1.3). Ifq(z) is a dominant of (1.3) and

q(z)≺q(z)for all dominants of (1.3), then we say thatq(z) is thebest dominant of the differential subordination (1.3).

Lemma1.1[6]. Letq(z)be univalent in the unit discU, and letθ(ω)andφ(ω)be analytic in a domainDcontainingq(U), withφ(ω)≠0whenω∈q(U). SetQ(z)=

zq_{(z)φ(q(z)), h(z)=θ(q(z))+Q(z), and suppose that} (i) Q(z)is starlike in the unit discU,

(ii) Re{z(h_{(z)/Q(z))} =Re{θ(q(z))/φ(q(z))+z(Q(z)/Q(z))}>0, z∈U.}

Ifp(z)is analytic inU, withp(0)=q(0),p(U)⊆Dand

θp(z)+zp_{(z)φp(z)≺θq(z)+zq(z)φq(z)=h(z) (1.4)}

thenp(z)≺q(z),andq(z)is the best dominant of (1.4).

Even more we need the following lemma, which in more general form is due to Hallenbeck and Ruscheweyh [3].

Lemma1.2[3]. LetG(z)be a convex univalent inU,G(0)=1. LetF(z)be analytic inU,F(0)=1and letF(z)≺G(z)inU. Then for alln∈N0

(n+1)z−n−1z 0t

n_{F(t)dt≺(n+1)z}−n−1z 0t

n_{G(t)dt. (1.5)}

2. Main results and consequences. In this part, we use Lemmas 1.1 and 1.2 to obtain some conditions forf (z)f_(z)/f2_{(z)which lead to starlikeness.}

Theorem2.1. Iff∈Aand

f (z)f_(z)

f2_(z) ≺2− 2

(1−z)2=h(z) (2.1)

thenf∈S∗_.

Proof. We choosep(z)=z(f_{(z)/f (z)); q(z)=(1−z)/(1+z); φ(ω)=1/ω}2_; θ(ω)=1−(1/ω). Thenq(z)is univalent in U; θ(ω) and φ(ω)are analytic with domainD=C\{0}which containsq(U)= {z: Re(z) >0}and φ(ω)≠0 whenω∈

q(U). Further

Q(z)=zq_{(z)φq(z)= −} 2z

(1−z)2 (2.2)

is starlike inU, and for the function

h(z)=θq(z)+Q(z)=2z(z−2)

(1−z)2 =2− 2

(1−z)2 (2.3)

we have

Re

zh_Q(z)(z)

=Re

₂

1−z

>0, z∈U. (2.4)

Also,pis analytic inU,p(0)=q(0)=1 andp(U)⊂Dbecause 0∈p(U). Therefore the conditions of Lemma 1.1 are satisfied and we obtain that if

θp(z)+zp_(z)φp(z)=f (z)f(z)

f2_(z) ≺2− 2

then

zf_(z)

f (z) =p(z)≺q(z)= 1−z

1+z, (2.6)

i.e.,f∈S∗_.

Example2.2. The functionf (z)=z−z2_{/2 belongs to the classAandf (z)f(z)/}

f2_{(z)=1/2−(1−z)}2_{/2 is subordinated to 2−2/(1−z)}2_{. So, from Theorem 2.1} f∈S∗_{. Obtaining starlikeness fromzf(z)/f (z)=(2−2z)/(2−z) needs one step} more.

Corollary2.3. Letf∈A.

(i) LetD= {z: Rez <1.5} ∪ {z: Rez≥1.5, |Imz|>√−3+2Rez}. Iff (z)f_(z)/

f2_{(z)∈D,z∈U, thenf∈S}∗_;

(ii) ifRe{f (z)f_(z)/f2_{(z)}<3/2, z∈U, thenf∈S}∗_; (iii) if|f (z)f_(z)/f2_{(z)|<3/2, z∈U, thenf∈S}∗_.

Proof. (i) We have thatf (z)f_(z)/f2_{(z)andh(z)defined by (2.1) are analytic} inU;f (0)f_(0)/f2_{(0)=h(0)=0 andh(z)is univalent inU(it is one to one mapping} because only one of the points 1+2/(2−ω)is inU). So, we get that (2.1) is equivalent with

f (z)f_(z)

f2_(z) ∈h(U), z∈U, (2.7)

and it is enough to prove thath(U)=D. After some transformations we obtain _h

eiθ₋₂₌ 1

2sin2_θ/2, arg h

eiθ_{−2= −θ, (2.8)}

i.e.,

Re heiθ₋₂₌1 2

ctg2θ 2−1

, Im heiθ_{= −ctg}θ

2. (2.9)

So

Im heiθ_{= ±−3+2Rehe}iθ _(2.10) and because ofh(0)=0<3/2 we can say thath(U)=D. Parts (ii) and (iii) follow directly from (i).

Example2.4. The functionf (z)=1−e−z_{is inAand the real part off (z)f(z)/}

f2_(z)=1−ez _{is smaller than 3/2 for all z∈U. Sof (z) is starlike according to} Corollary 2.3(ii). It have been more complicated to realize it from zf_{(z)/f (z) =}

z/(ez_−1).

Now, using Lemma 1.2 we prove a theorem which we used to improve the results from Corollary 2.3(ii) and (iii) and to obtain some other results.

Theorem2.5. Letf∈A. Iff (z)f_(z)/f2_{(z)≺h(z), h(0)=0andh(z)is a}

con-vex univalent inUthen

f (z) zf_(z)≺1−

1 z

z

Proof. Let F(z)=(f (z)/f_(z))

=1−f (z)f_(z)/f2_{(z) and G(z) =1−h(z),} z∈U. ThenG(z)is a convex univalent inU,G(0)=1;F(z)is analytic inU,F(0)=1. Further we have that

1−f (z)f_f2(Z)(z)=F(z)≺G(z)=1−h(z). (2.12) Therefore the conditions of Lemma 1.2 are satisfied and forn=0 we obtain

1 z

z

0F(t)dt≺ 1 z

z

0G(t)dt. (2.13)

If we apply the definitions ofF(z)andG(z)in the result above and use the following fact which is true becauseF(z)is analytic

z

0

f (t) f_(t)

dt=_ff (z)_(z)−_ff (₍0₀)₎=_ff (z)_(z), (2.14)

we obtain that

f (z) zf_(z)≺

1 z

z

0

1−h(t)dt=1−1

z z

0h(t)dt. (2.15)

Remark2.6. Ifh(z)is convex, from [4], 1−(1/z)0zh(t)dtis also convex. In the following corollaries, we deliver some interesting results using Theorem 2.5. Corollary2.7. Letf∈A. If|f (z)f_(z)/f2_{(z)|<2thenf∈S}∗_.

Proof. From|f (z)f_(z)/f2_{(z)|<2,z∈U, becauseh(z)=2zis univalent and} f (0)f_(0)/f2_{(0)=h(0)=0 we get that}

f (z)f_(z)

f2_(z) ≺2z=h(z). (2.16)

Further,h(z)is convex, so the conditions from Theorem 2.5 are satisfied, and we obtain

f (z) zf_(z)≺1−

1 z

z

0h(t)dt=1−z, (2.17)

i.e.,

Re f (z) zf_(z)

>0. (2.18)

Because of that Re{zf_{(z)/f (z)}>0, i.e.,f∈S}∗_.

Remark2.8. The result from Corollary 2.7 is the same as in [7] (Theorem 1, for a=0 andb= −1) and it is better than the result from Corollary 2.3(iii).

Example2.9. The same function as in Example 2.4,f (z)=1−e−z_{, can be used to} illustrate Corollary 2.7:

f (z)f_f2_(z)(z)

Corollary2.10. Letf∈A.

(i) Iff (z)f_(z)/f2_{(z)≺2αz/(1+z)=h(z), 0< α≤1/2(1−ln2), thenf∈S}∗_. (ii) IfRe{f (z)f_(z)/f2_{(z)}<1/2(1−ln2)=1.629445..., z∈U, thenf∈S}∗_. Proof. (i) Fromh(0)=0 andh(z) is a convex function in the unit discU, by Theorem 2.5 we get that

f (z) zf_(z)≺1−

1 z

z

0h(t)dt=1−2α+2α

ln(1+z)

z =g(z). (2.20)

Now, from

Re g(z)=1−2α+_|2_zα_|2xln|1+z|+yarg(1+z),

Im g(z)=_|2_zα_|2xarg(1+z)−yln|1+z|,

(2.21)

wherez=x+iy, it follows thatg(U)is symmetric with respect to thex-axis. It is also convex (Remark 2.6) and so

Re g(z)>min g(1),g(−1)=g(1)=1−2α+2αln2>0, z∈U. (2.22)

Thus, from f (z)/zf_{(z) ≺ g(z) we get that Re{f (z)/zf(z)} > 0, z ∈ U and} Re{z(f_{(z)/f (z))}>0,z∈U, i.e.,f∈S}∗_.

(ii) f (z)f_(z)/f2_{(z) is analytic in the unit disc U, h(z) is univalent in U and} f (0)f_(0)/f2_{(0)=h(0)=0. Therefore the condition from (i)}

f (z)f_(z)

f2_(z) ≺ 2αz

1+z=h(z) (2.23)

is equivalent with

f (z)f_(z)

f2_(z) ∈h(U), z∈U. (2.24)

Now, from Re heiθ_{=αandh(0)=0< αwe get thath(z)maps the unit discU} into the half plane with real part less thanα. So the condition from (i) is equivalent with

Re

_{f (z)f(z)} f2_(z)

< α, z∈U. (2.25)

If we putα=1/2(1−ln2)here, using (i) we obtain the statement of (ii).

Remark2.11. Because 1/2(1−ln2)=1.629445···>1.5, the result from Corollary 2.10(ii) is better than the result from the Corollary 2.3(ii).

Example2.12. Forf (z)=(1−e−2z_{)/2 we have thatf∈Aandf (z)f(z)/f}2_(z)= 1−e2z_{. Further forz=e}iθ_{we get}

with maximum value 1.603838... which it attains for θ=1.246054...,i.e., for the solution of the equation

θ+2sinθ=π. (2.27)

So from Corollary 2.10(ii) we obtain thatf (z)is starlike. Starlikeness off (z)could not have been derived using Corollary 2.3. Also, because forz=1

f (z)f_f2_(z)(z)

= |1−e2z_{|>2, (2.28)} we cannot use Corollary 2.7.

In the following corollary, we see what is happening iff (z)f_(z)/f2_{(z),z∈U, is} in the half plane right from 1/2(1−ln2).

Corollary2.13. Letf∈A.

(i) Iff (z)f_(z)/f2_{(z)≺ −ln(1+αz)=h(z),0< α≤1,thenf∈S}∗_;

(ii) If Re{f (z)f_(z)/f2_{(z)} ≥ a > −ln2 = −0.6931... and |Im{f (z)f(z)/}

f2_{(z)}|<arccos1/(2e}a_{), z∈U,thenf∈S}∗_.

Proof. (i)h(0)=0 andh(z) is a univalent function in the unit discU because h(z)is analytic inUand it is one to one mapping. Fromα≤1 we get that

Re1+zh(z) h_(z)

=Re 1

1+αz

>0, z∈U, (2.29)

i.e.,h(z)is a convex function in the unit discU. Therefore from Theorem 2.5 we obtain f (z)

zf_(z)≺1− 1 z

z

0h(t)dt=

1+_αz1

ln(1+αz)=g(z). (2.30)

Now, g(U) is symmetric with respect to the x-axis and g(z) is a convex function (Remark 2.6). So forz∈U

Re g(z)>min g(1),g(−1)

=min

1+_α1

ln(1+α),

1−_α1

ln(1−α)

≥0 (2.31)

and fromf (z)/zf_{(z)≺g(z)we get that Re{f (z)/zf(z)}>0, i.e., Re{zf(z)/f (z)}}

>0,z∈U, andf∈S∗_.

(ii) f (z)f_(z)/f2_{(z) is analytic in the unit disc U, h(z) is univalent in U and} f (0)f_(0)/f2_{(0)=h(0)=0. Therefore the condition from (i), forα=1}

f (z)f_(z)

f2_(z) ≺ −ln(1+z)=h1(z) (2.32) is equivalent to

f (z)f_(z)

f2_(z) ∈h1(U), z∈U. (2.33)

Further, Re h1eiθ_{= −ln2cos(θ/2)and Im h1e}iθ_{= −arg1+e}iθ_{= −θ/2. So,} Re h1eiθ_{≥afor|θ| ≥2arccos1/(2e}a_{), and for suchθwe get thatIm h1e}iθ₌

Example2.14. The use of Corollary 2.13 can be illustrated with the functionf (z)= ln(1+z). It belongs to the class A and f (z)f_(z)/f2_{(z)= −ln(1+z), so from} Corollary 2.13(i), fora=1, we get thatf∈S∗_{. The starlikeness is not obvious from}

zf_{(z)/f (z)=z/((1+z)ln(1+z)).}

References

[1] A. W. Goodman,Univalent Functions. Vol. I, Mariner Publishing Co. Inc., Tampa, Florida, 1983. MR 85j:30035a.

[2] , Univalent Functions. Vol. II, Mariner Publishing Co. Inc., Tampa, Florida, 1983. MR 85j:30035b.

[3] D. J. Hallenbeck and S. Ruscheweyh,Subordination by convex functions, Proc. Amer. Math. Soc.52(1975), 191–195. MR 51#10603. Zbl 311.30010.

[4] R. J. Libera,Some classes of regular univalent functions, Proc. Amer. Math. Soc.16(1965), 755–758. MR 31#2389. Zbl 158.07702.

[5] S. S. Miller and P. T. Mocanu,Differential subordinations and univalent functions, Michigan Math. J.28(1981), no. 2, 157–172. MR 83c:30017. Zbl 456.30022.

[6] , On some classes of first-order differential subordinations, Michigan Math. J.32 (1985), no. 2, 185–195. MR 86h:30046. Zbl 575.30019.

[7] M. Obradovi´c and S. Owa,On certain properties for some classes of starlike functions, J. Math. Anal. Appl.145(1990), no. 2, 357–364. MR 91d:30015. Zbl 707.30009.

Tuneski: Department of Mathematics and Computer Sciences, Faculty of Mechanical Engineering, University of “Sv. Kiril i Metodij”, Karpos II b.b.91000Skopje, Macedonia